Calculating limit of function To find limit of $\lim_{x\to 0}\frac {\cos(\sin x) - \cos x}{x^4} $.
I differentiated it using L Hospital's rule. I got
$$\frac{-\sin(\sin x)\cos x + \sin x}{4x^3}\text{.}$$ I divided and multiplied by $\sin x$.
Since $\lim_{x\to 0}\frac{\sin x}{x} = 1$, thus I got
$\frac{1-\cos x}{4x^2}$.On applying standard limits, I get answer $\frac18$. But correct answer is $\frac16$. Please help. 
 A: Without L'Hôpital
The expansions $\sin(x) = x - \frac{x^3}{6} + O(x^5)$ and $\cos(u) = 1 - \frac{u^2}{2} + \frac{u^4}{24} + O(u^6)$ combine joyfully to give
$$
\cos(\sin x) - \cos(x) = \left(1 - \frac{x^2}{2} + \frac{5x^4}{24}\right) - \left(1-\frac{x^2}{2} + \frac{x^4}{24}\right) + O(x^5)
$$
so finally,
$$
\lim_{x\to 0} \frac{\cos(\sin x) - \cos(x)}{x^4} = \frac{5}{24}-\frac{1}{24} = \frac{1}{6}.
$$
A: Using Prosthaphaeresis Formulas,
$$\cos(\sin x)-\cos x=2\sin\frac{x-\sin x}2\sin\frac{x+\sin x}2$$
So, $$\frac{\cos(\sin x)-\cos x}{x^4}=2\frac{\sin\frac{x-\sin x}2}{\frac{x-\sin x}2}\frac{\sin\frac{x+\sin x}2}{\frac{x+\sin x}2}\cdot\frac{x-\sin x}{x^3}\cdot\frac{x+\sin x}x\cdot\frac14$$
We know, $\lim_{h\to0}\frac{\sin h}h=1$
$$\text{Apply L'Hospital's Rule on  }\lim_{x\to0}\frac{x-\sin x}{x^3}$$
$$\text{and we get }\lim_{x\to0}\frac{x+\sin x}x=1+\lim_{x\to0}\frac{\sin x}x$$
A: One approach to problems like this is to replace the numerator and denominator by functions that have the same value, derivative, second derivative, etc. (up to some point) as the originals at the limit point. For instance, to handle $\sin(x)/x$, you replace $\sin(x)$ with $f(x) = x - x^3/3!$, which agrees with $\sin$ to 3rd order at $x = 0$,  i.e., $\sin(0) = f(0)$; $\sin'(0) = f'(0)$; $\sin''(x) = f''(0)$; and $sin^{(3)}(0) = f^{(3)}(0)$. So then you look at $\lim \frac{x - x^3/6}{x} = \lim 1 - x^2/6 = 1$, and you're done. 
In the case of this problem, you can replace $\sin(x)$ with $x - x^3/6$ and $\cos(x)$ with $1 - x^2/2$, so that 
$\cos(\sin(x)) - \cos(x))$ 
becomes
$1 - \frac{(x - x^3/6)^2}{2} - (1 - x^2/2)$
When you simplify that a bit, you'll get to the answer. 
